Question

Be sure to answer all parts.

Find the pH of the equivalence point and the volume (mL) of 0.125 M HCl needed to reach the equivalence point in the titration of 65.5 mL of 0.234 M NH₃.

Volume =

mL HCl

pH =

Answer 1

1)

find the volume of HBr used to reach equivalence point

M(NH3)*V(NH3) =M(HBr)*V(HBr)

0.234 M *65.5 mL = 0.125M *V(HBr)

V(HBr) = 122.616 mL

Answer: 123 mL

2)

Given:

M(HBr) = 0.125 M

V(HBr) = 122.616 mL

M(NH3) = 0.234 M

V(NH3) = 65.5 mL

mol(HBr) = M(HBr) * V(HBr)

mol(HBr) = 0.125 M * 122.616 mL = 15.327 mmol

mol(NH3) = M(NH3) * V(NH3)

mol(NH3) = 0.234 M * 65.5 mL = 15.327 mmol

We have:

mol(HBr) = 15.327 mmol

mol(NH3) = 15.327 mmol

15.327 mmol of both will react to form NH4+ and H2O

NH4+ here is strong acid

NH4+ formed = 15.327 mmol

Volume of Solution = 122.616 + 65.5 = 188.116 mL

Ka of NH4+ = Kw/Kb = 1.0E-14/1.8E-5 = 5.556*10^-10

concentration ofNH4+,c = 15.327 mmol/188.116 mL = 0.0815 M

NH4+ + H2O -----> NH3 + H+

8.148*10^-2 0 0

8.148*10^-2-x x x

Ka = [H+][NH3]/[NH4+]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((5.556*10^-10)*8.148*10^-2) = 6.728*10^-6

since c is much greater than x, our assumption is correct

so, x = 6.728*10^-6 M

[H+] = x = 6.728*10^-6 M

use:

pH = -log [H+]

= -log (6.728*10^-6)

= 5.1721

Answer: 5.17